 Welcome to class 31 on topics in power electronics and distributed generation. In the last class we have been looking previously, we have been looking at the selection of DC bus capacitors in a inverter. So, today we will look at selection of the power semiconductor devices in a inverter. So, if you look at the what we have been studying so far in our switching models we have been considering an ideal switch. If you look at an ideal switch ideal switch will block infinite voltage and it will not have any leakage current when it is blocking the voltage. The ideal switch can actually carry an infinite amount of current without having a substantial on state drop. So, there is no on state drop in the ideal switch whereas in a practical switch you are going to have a finite voltage drop and there will be conduction losses because of the on state drop in the semiconductor device. If you look at the switching transition times in the ideal switch we are considering it going from on to off and off to on on a instantaneous basis in the ideal switch whereas in a practical switch you have finite transition times between your on state and the off state. So, if you look at a practical switch so practical switches are the most basic is the diode, you have thyristors, you have MOSFETs, IGBTs, GTOs etcetera. For many applications in the distributed generation range we could consider the power levels of a few tens to hundreds of watts to a few tens to hundreds of kilowatts. You have commonly devices such as diodes, IGBTs, MOSFETs etcetera. So, we will consider a IGBT and a diode as a standard practical switch when we look at the semiconductor device. So, if you look at a diode it is a uncontrolled switch. So, when the external voltage from anode to cathode is positive it would conduct current when you reverse the voltage if when you apply a negative voltage across the terminals it would block conduction it would not conduct current. If you look at a control switch you have an IGBT or a diode IGBT stands for insulated gate bipolar transistor or you have a MOSFET, metal oxide field effect transistor. These are gate control devices. So, depending on the voltage that you externally apply at the gate you can control the switch to be in on state or in off state. If you look at the breakdown voltage of these devices you have a finite breakdown voltage. We are looking at examples where we have DC bus of the order of 600 to 800 volts or 400 volts. So, you might have common semiconductor blocking voltages or 600 volts rated semiconductors or 1200 or 1700 volt rated semiconductors these are common blocking voltages 400 volts. You also have medium voltage IGBTs which can go up to 3.3 kV, 6.6 kV etcetera, but only for DG applications you would be using IGBTs that can be used in low voltage applications. And you are if you are looking at the leakage current the leakage current is actually a function of temperature you heat a semiconductor device the leakage current typically would increase you are talking of currents in the range of micro amps to nano amps depending on the rated current of the semiconductor. So, higher rated current semiconductor would have higher leakage current compared to one at the lower rated current because physically the cross sectional area of the chip would be lesser for a lower rated current device. And it is also dependent on temperature. If you look at the finite on state voltage drop. So, we are talking about 1 to 1.5 volts for typically for diodes you are talking about again 1.5 volts to 2 volts for IGBTs. So, for MOSFETs depending on your on state resistance you might have voltages in the range of the order of 1 volt depending on the current level and the on state resistance. If you look at your switching transition times today switches are able to turn on and turn off in the range of 10s of nanoseconds. So, of the order of 10s of nanoseconds. So, again depending on the rating of the device larger rating device might be slower you have finite switch transition times. So, you are able to switch at 10s to 100s of kilohertz with the semiconductor devices that are available today. So, if you compare a ideal switch because you had no leakage current no 0 on state drop you had 0 switching time the switching the power loss in the ideal switch is 0. Whereas, because of all these reasons in the practical switch you would have finite power loss. So, the implication of all these factors is p loss in the semiconductor is finite and that has important implications on how you operate the power converter. So, if you look at the on state loss this is commonly referred to as a conduction loss and depending on the on state voltage and the current that is being carried by the semiconductor you would have participation in the device which can be a significant factor especially if you look at the overall loss in a device. So, ideally you would like to have a device with low on state voltage to actually minimize conduction loss. If you look at the off state power loss this corresponds to due to leakage current. So, in the conduction loss if you have a on state voltage drop of the order of 1 volt and if you are carrying currents of the order of 10 amps 100 amps etcetera you are talking about order of 10s of watts of participation in the in case of leakage current you are talking about off state voltages of the order of 800 volts may be 1000 volts and if you are talking about micro amps of leakage current or nano amps. So, you are talking about participations of milliwatts. So, compared to the conduction drop the off state power loss may be neglected especially when you are looking at low voltage applications. And typically when we are referring to low voltage in from a power systems perspective we are talking about voltages less than 1000 volts. If you look at your switching loss you have switching transitions from off to on turn on and on to off. So, this would correspond to the switching loss and because of the finite loss that occurs in a device at each switching event if you now operate at a higher switching frequency your switching loss will scale with your switching frequency. So, one limitation of how fast you could switch a device is that your power loss due to the switching loss factor would increase to such an extent which might limit how fast you can switch a device. And you could say what is the final big concern for doing a detailed evaluation of power loss in a semiconductor device in a power converter. The primary implication is to ensure that your maximum temperature within your semiconductor does not exceed critical values and to ensure that your semiconductor devices stay cool during the operation of your power converter. And your overall losses have to sum through all these factors to when you are actually evaluating your overall loss to look at the junction temperatures. So, if you look at example of structure of a semiconductor device. So, here what is shown is a schematic of cross section of a IGBT semiconductor chip. This is actually a vertical double diffuse because you have diffusion for N plus and P layers within the semiconductor. You have the N plus would be connected to the emitter. So, you have metallization which is shown in the upper blue curve and what is shown within the red curve is the gate oxide. So, the gate metallization is what is shown in the magenta color. So, you have connections to the gate oxide, within the gate oxide you have the gate. So, you have the aluminum which is used as a gate connection to the IGBT. So, you could apply a gate voltage with respect to the emitter. If you look at the region where the gate is located it is located between the emitter and the N minus region. So, you have a channel that can be where you can induce charges by applying appropriate voltage. For example, if you apply a positive voltage between the gate and emitter you would get positive charges in the gate metallization which would induce negative charges just below in the P region of the semiconductor. So, this N minus region is called a drift region which is primarily responsible for the blocking voltage. So, if you need larger blocking voltage the drift region would be wider and the junction which actually blocks the voltage is this particular P N junction. So, if you apply a positive voltage at the collector you do not have gate voltage. So, you apply positive to the N minus region and negative to the P. So, it is a reverse biased junction. So, you have depletion regions that are formed across this N minus P region because this is lightly doped the depletion region extends well down into the structure. So, essentially this particular dimension and the doping level of this region determines what the rated voltage the breakdown voltage of the devices. The P plus region at the bottom is what is connected to the collector of the device. So, if you when you are passing current positive current from your collector to the emitter it essentially flows from the P plus to the N plus out and out through the emitter. So, if you apply a voltage if you apply a positive voltage VGE between your gate and the emitter essentially you will induce charges in this particular region and you can get essentially electrons to flow in this particular direction and holes to flow back. You also have a P N P transistor over here which would give you enhancements in terms of the conductivity. The P N junction would inject minority carriers into the N minus region which would give you conductivity modulation which reduces the on state voltage drop. So, if VGE is greater than the threshold voltage of that particular gate oxide and the semiconductor. So, if it is greater than the threshold voltage essentially your device would start conducting current and if your gate voltage is below the threshold voltage it would essentially block conduction. If you look at the structure of the IGBT you have a P N P transistor you also have a N P N transistor in early IGBTs people used to worry about the problem of latching of IGBTs because you have essentially a P N P N thyristor structure. If you look at IGBTs that are available today this is no longer a problem. We have IGBTs with a rugged square safe operating area which can be used without problems of latching. If you look at the IGBTs operating region in a power converter you are either going to be using it in the off state. So, this would correspond to applying a voltage and blocking current or you would be using it in the on state. So, you are switching between the on and off conditions if you look at if you are in the off state as long as you ensure that your voltage does not exceed the breakdown voltage. If you if your voltage exceeds the breakdown voltage then essentially you would have uncontrolled conduction. So, your voltage rating is less than v breakdown of the junction. So, the breakdown of the junction depends on the geometry, the temperature of the of the device etcetera and for 800 volts DC bus we would commonly use a 1200 volt IGBT. So, if you look at why you need the margin of 400 volts going from 800 to 1200 volt device you can look at a typical IGBT module. You would always have interconnection inductance say if you have a L stray of the order of say 10 nanohenries to 30 nanohenries. So, this you can get 30 nanohenries even in few within less than 10 centimeters of interconnection. You have your actual device voltage would be is v DC plus L stray into di by dt. So, for example, when you are switching off the device you will end up with a di by dt which applies a voltage across the L stray. So, if you look at your di dt's that are of present day switches you are talking about 1 to 10 kilo amps per microsecond. So, if you look at at 10 kilo amps per microsecond you will end up with a fair amount of voltage drop a voltage that is being applied across the device. So, whether it is 1 or whether it is 10 is also a function of what type of gate resistances you would use if you use a smaller value of gate resistance your di dt's can be larger. If you use a large value of gate resistance your di dt's would be smaller, but then your switching losses would also increase. So, depends on your R gate. If you are then talking about say something like 10 kilo amps per microsecond and say you are talking about L stray of 10 nano henries you are talking about 100 volts to 300 volts even with 30 nano henries you would get 300 volts drop across your stray inductance which can then apply a substantial voltage across your semiconductor device. So, you definitely need a margin above what is available on the DC bus. So, it is critical for a designer to ensure that the inductance of your loop which includes your switching module is extremely small. So, people look at parallel plate structures if you are using capacitors you need to ensure that you have low ESL parts available in parallel to take up fast switching transitions even a physical module might have internal inductance of the order of 10 nano henries. So, to ensure that your devices stay safe without breaking down you need to coordinate your gate resistance selection with tight geometric design of your DC bus. So, if your voltage exceeds your rate breakdown voltage then you have avalanche breakdown. So, essentially the electric field has become large enough that between two collisions of the carriers it speeds up sufficiently to actually generate additional carriers. So, you lose control of the gate and you end up with a large amount of current that can actually flow through the device in an uncontrolled manner. So, implication is now you can end up with a large participation and eventually leading to device failure. So, it is important to ensure that you are always operating your semiconductor below its rated voltage. If you look at the other aspect of voltage rating. So, if you look at we have looked at the voltage that can be applied say from the collector to emitter the other voltage rating is related to how much voltage you can apply between the gate and emitter. So, if you typical voltages that are used at the gate emitter junction this of the order of 15 volts or 15 volts is a common voltage. If you apply excessive voltage you can actually damage the oxide also if you apply higher level of gate voltage essentially with increasing gate voltage your saturation current within your device is going up. So, at a higher voltage you will end up with higher saturation currents which is what would flow in case there is a short circuit in the particular device. So, your device can get rapidly damaged if you apply excessive gate voltage. If you apply a gate voltage which is low instead of having a on state voltage over here you might end up with a on state voltage over here. So, you might end up with higher conduction losses if the gate voltage is low commonly used gate voltage is 15 volts. So, 15 volts is a typical number also if you apply a excessive Vg you can end up damaging a gate oxide which would then damage your device. So, if you then look at what could be the next rating factor which people would typically consider is the current rating of a semiconductor. So, when you look at the current rating of the semiconductor which is actually provided prominently in the data sheet we need to remember that the current rating is only a indicative number it is does not specify the actual application in which your semiconductor is going to be used. So, if you look at what your current rating is it is determined as I rated and your on state voltage when your device is on is VCE on and essentially the product of this is your conduction loss your point conduction loss power and essentially if you look at what your Tj junction maximum temperature is you might have semiconductors which are rated for 150 degrees some might be rated for 125 degrees. So, you look at your Tj max and you look at your case temperature which is specified. So, people might specify your current rating at 25 degrees or at 70 degrees or some particular indicated temperature. So, this is essentially at room temperature or some particular operating temperature times your thermal resistance of your device your thermal resistance from junction to case we will talk more about the junction temperature shortly. So, essentially if you look at your rated current it is essentially having factors only of the conduction drop under specified measurement conditions. So, it is only a indicative number if you look at your actual operating current rating with your switching action and the associated switching losses your actual current capability of your semiconductor would be less than your rated current. So, however you need to also keep in mind that the current rating is also indicative not just of the semiconductor it is also indicative of the current carrying capability of your connectors your terminals your wire bonds etcetera within your device. So, it is not just of the semiconductor it is actually of your entire package and if you operate above your current rating then it has to be for extremely short durations. For example, when you have a short circuit you if you operate for just a few less than 10 microseconds you might be able to save your device. But if you continuously operate above the rated current you can actually damage sub components within your semiconductor package. So, the implications of operating above your current rating is that your devices would heat up you can immediately see from this expression for how the rating is being defined that is actually linked to your temperature. So, if you look at typical power module. So, if you look at a typical power module you would have what is shown within this green box you would have multiple chips corresponding to switches diodes. Here what is shown is a leg you would have a switch for the upper switch in the leg lower switch in the leg diodes D 1 and D 2. In turn if you have higher current rated modules each of these red squares might actually contain multiple chips which are operated in parallel. So, you might have dozens of semiconductor chips which are packaged within a semiconductor module. So, if you look at the to evaluate the thermal rating of the semiconductor module one have to look at all your components your transistor diode in parallel which may be in parallel. If you would have solder joints or bracing joints wire bonds you would have ceramic materials for isolation. Typically the chips are soldered on to a ceramic and the ceramic is then soldered on to a base plate and is a base plate which conducts heat out of the module. You have you might have internal resistors say you might have resistors for gate paralleling. You might have sensors especially if you have newer modules come integrated with multiple functions people call them intelligent power modules. You might have sensors you might have temperature sensor you might have current sensors you might have DC bus voltage sensing. Also the IPMs intelligent power modules might also have integrated gate drive circuits. So, one has to look at the limitations of all the components within the particular package and if you look at a critical aspect is to look at how much participation is happening in the transistor and the diode which is actually carrying the largest level of power within the overall module structure. So, again what is the issue of the higher temperature is that at higher temperature your semiconductor the carriers the carriers within the semiconductor becomes excessive and once the excess carriers the value goes beyond what is there for your background doping and the of your semiconductor then essentially the capability of your junctions to block goes away and you would have uncontrolled conduction. So, typically if you look at the temperatures at which your semiconductor material would have such large amount of excess carriers would be of the order of say 300 to say 400 degree centigrade. If you look at the actual semiconductor chips they are highly inhomogeneous. So, people define essentially a virtual temperature which is what people commonly refer to as the junction temperature. So, and you have a number of non-homogeneities you are talking about non-symmetric cooling. So, the upper layer of the semiconductors would have lower cooling than the lower levels of the semiconductor. If you look at the wire bonding connections to the semiconductor it is not uniform across the surface they are at discrete points. So, if you look at the actual chip the edges might be at much cooler than the center of the chip. So, you have connections for the gate which introduces non-homogeneity in the participation across the chip you have multiple chips. So, the chips that are at the edge would get better cooling than chips that are surrounded by many other chips which are themselves dissipating power. You also have passivation materials around the chip which makes the participation within the chip non-uniform. So, when you are talking about temperature there is a significant difference between your actual junction temperatures to what people refer to as virtual junction temperature which one would like to keep below 125 or 150 degree centigrade. So, you can see that clearly you are having a factor of two factor of safety between your actual junction temperatures which might actually be very difficult to measure to measurements on the surface of the chips which might be lower than what is actually happening at the junction to a evaluated junction temperature virtual junction temperature which is kept of the order of 125 to 150. The assumption is that if you keep your virtual junction temperature to less than 150 then the overall system does not go into a condition where your excess carrier background carriers levels become too high that you go into uncontrolled conduction and excessive leakage currents. So, the junction temperature Tj need to be evaluated not just for your transistor but also for your diode. So, one would evaluate what losses are happening in your IGBT switches and what losses are happening in your diode you look at it separately from your conduction perspective conduction loss perspective and your switching loss perspective. So, to look at your conduction loss essentially a model for your conduction loss might involve the detailed physics of your junctions which would give you the exact expression for what your voltage drop would be at a given level of current when a device is on or it can be simplified as shown in this figure to a voltage VCEO on state plus what is shown over here is the reciprocal which is the reciprocal of R on on state voltage of on state resistance of your semiconductor device. So, if you look at your VCE sat your on state saturation voltage and if you look at essentially what your IC is whatever current is going out through the output AC of your inverter leg is essentially going to flow as the collector current and essentially your expression for your on state voltage is can be modeled as a constant voltage plus a IR term drop term. Both these values VCEO and R on are a function of temperature. So, if you change your temperature your on state voltage would actually change if you look at a power semiconductor device with positive temperature coefficient of on state drop your VCE sat would increase with temperature. So, if you essentially what that implies would is that a hotter device would actually carry lesser current because especially when two devices are put in parallel a hotter device would have a larger drop which means that the other device which is in parallel would be forced to take up the current. So, and essentially what this implies is that you will not have a situation of thermal runaway where if you had a hotter device which whose drop would reduce then it would actually draw more current and eventually that particular device which is hottest would end up carrying all the current and have the maximum power dissipation. So, if you look at the power dissipation during the conduction we have the conduction in your transistor is essentially VCE sat into IC and this is equal to VCEO times IC plus R on into IC square. So, if you look at the terms corresponding to the power conduction loss this term corresponds the first term with VCEO corresponds to the power loss due to the average current that is flowing during say a switching duration. So, and the second term corresponds to a I square R type of situation which would correspond to a RMS heating effect. You could write a similar expression for your diode your P conduction of your diode is V diode saturation voltage times I d again can be modeled as V d 0 into IC which would be essentially your I out plus R d on to I out square. So, if you look at the switching action between your in your power converter which is all happening your transferring power from a voltage source side you might have some strain inductance, but essentially you have a voltage source on one side and you have a filter which would represent a filter inductor which would correspond to a current source. So, essentially the switching action of this leg corresponds to power transfer between a voltage source and a current source. So, you could look at then the next component of the power loss which is essentially your switching loss and if you look at your switching action you have essentially your turn on and you have your turn off. So, if you look at the sequence of events in a typical switching transition first you would have a your gate which is actually be triggered high which essentially your gate voltage would be the commanded to go to plus 15 volts. Once your gate voltage reaches the threshold voltage then essentially your current starts rising and once the current rises above your actual output current you have essentially your recovery of your reverse your freewheeling diode. So, essentially if you look at a situation where your current is positive and you are trying to say turn on switch S 1. So, originally your diode D 2 was conducting this particular positive current. So, during turn on your current starts rising once the current rises above your I out then essentially your excess carriers which are stored within your diode is what is swept out by this particular current which leads to spike in this red wave form over here. So, this corresponds to your reverse recovery. So, once the reverse recovery phenomena of the diode is completed when the essentially your voltage starts dropping down your original voltage which was your DC voltage which the switch was blocking would now turn come down to your on state saturation voltage. So, this level over here is your V C E SAT which is modeled when you look at your conduction drop. So, if you look at the period of your switching action this is starting from when your current starts rising to your voltage starts reaching quite close to your saturation voltage. So, this particular duration is when there can be a significant amount of power dissipation within your device. So, if you look at your instantaneous power dissipation your P instantaneous essentially it is a product of V C E times I C. So, you can see at the peak point over here essentially you are multiplying your V DC times the current which is essentially your I out plus your reverse recovery current. So, if you are talking about so, this is proportional your peak value of your participation is proportional to your DC voltage and it is proportional to your I out. So, if you are talking about say DC bus voltage of say 800 volts. So, this and say you are talking about current levels of the order of say 100 amps then you are talking about participation of the order of 80 kilowatts. So, you can see that there can be a significant participation during the switching action and your ideal switch had a participation of 0 and you can see that during a inductive turn on and same case for the turn off you can have fairly significant participation and the way to ensure that this participation leads to lesser overall power dissipation is by ensuring that this duration during which this occurs is extremely small. So, this is what we want to do by keeping your gate resistances to be quite small and ensuring that even if there is a large participation it is occurring over a very small interval. So, as to ensure that your energy that is dissipated is extremely small. So, if you look at your E on energy during the switching action that would be your integral from 0 to T on and your actual energy dissipation is a function of your junction temperature, your gate resistance and your gate voltage that is being applied and essentially manufacturers provide data sheets which contain your turn on losses at different current levels for typical gate resistance that are suggested by the manufacturers. So, also your data sheet would contain your E on at I rated and at some nominal DC bus voltage and you might have again at a couple of temperatures you might have it at your room temperatures or your maximum junction temperatures. So, one could actually go by the explicit curves that are provided by the manufacturers or one could actually look at your actual operating condition might not be just at your rated condition at the specified DC nominal that is being used by the manufacturer. For example, the manufacturer might have specified your DC voltage at 700 volts you might be using it at 600 volts or may be 800 volts your rated current might be 100 amps, but you might be operating your converter anywhere between 0 amps to may be 50 amps. So, it is important to evaluate your switching energy loss under conditions which are different from what might be specified in the data sheets. So, one way to do it is by actually having curves which interpolate what the switching loss would be at your actual V dc and your actual collector current or your output current would be E on under rated conditions times V dc actual divided by V dc norm raised to a power k V and your actual output current divided by I rated the power of k I often people take k I to be equal to 1 and you could actually then derive expressions for your switching turn on loss over a fundamental cycle or your switching loss over a fundamental cycle. If you look at your turn off transient similar to the turn on in this particular case you would have your gate voltage being commanded to go off and some time later your collector voltage would start rising and once your collector voltage reaches your dc bus voltage then your output current starts falling and because your output current is falling with some large fairly significant d i d t you end up with a l d i by d t l straight d i by d t which applies a over voltage on the device and your current fall might be followed by a tail current the tail current would correspond to sweep out of carriers from within the i g b t. So, you can then again define a turn off duration of the switching device and if you look at the instantaneous participation in the device during turn off you could write expressions similar to what we did for turn on and look at what the losses would be during the turn off transient to. So, if you look at your turn off and this is a function again of your junction temperature your r g off and your v g e off and you could write expressions for to interpolate your off state switching loss for your transistor at decide dc bus voltage and your current. So, these expressions are actually given for the switch you could similarly write an expression for your loss switching loss of your diode and if you look at your switching loss of your diode the most important switching loss is the reverse recovery loss in the diode often the factors k i is taken as 1. So, that you could get a linear relationship between your output current and your loss to help obtain simplified expressions for the for the for their power loss. In case of reverse recovery it is typically not that linear because you can have significant reverse recovery even when your output current is close to 0 and your reverse recovery current may not change that much with loading as much as what your transistor switching loss terms do. So, once you have your switching loss you could then try to actually look at what is your overall loss within your switch and the diode you can write expressions for power loss in the switch and the diode and we will start with that particular exercise. So, if you look at the devices that would conduct when I out is positive essentially the devices that would conduct is switch s 1 and diode d 2 if it was negative if you had current flowing in say the opposite direction then switch s 2 and diode d 1 would be the devices that would conduct. So, depending on the polarity of the current you will have to actually look at what would be the device what would be the particular chip that would have switching loss and conduction loss. So, you would can write an expression for when switching loss is happening in particular switch for a diode and total that over a fundamental cycle to actually look at what happens over the AC cycle which would be used in typical d g systems. Thank you.